Solutions to the Dirac equation(Pauli–Dirac representation)

Dirac equation is given by

(iγµ∂µ −m)ψ = 0. (1)

To obtain solutions, we fix our convention (Pauli–Dirac representation for Cliffordalgebra) to the following one:

γ0 =

(1 00 −1

), γi =

(0 σi

−σi 0

). (2)

It is easy to check that these matrices satisfy the Clifford algebra {γµ, γν} = 2gµν .To find solutions to this equation, it is convenient to rewrite it in the form similarto that of Schrodinger equation,

ih∂

∂tψ = c(~α · ~p +mcβ)ψ, (3)

where ~p = −ih~∇ (to be distinguished with c-number ~p). Below, I take h = c = 1as usual. The matrices ~α and β are defined by

β = γ0, ~α = γ0~γ

(−σi 0

0 σi

). (4)

First, for a momentum

~p = p(sin θ cosφ, sin θ sinφ, cos θ),

we define two-component eigen-states of the matrix ~σ · ~p for later convenience:

χ+(~p) =

(cos θ

2

sin θ2eiφ

), (5)

χ−(~p) =

(− sin θ

2e−iφ

cos θ2

), (6)

which satisfy(~σ · ~p)χ±(~p) = ±pχ±(~p). (7)

Using χ±, we can write down solutions to the Dirac equation in a simple manner.

Positive energy solutions with momentum ~p have space and time dependenceψ±(x, t) = u±(p)e−iEt+i~p·~x. The subscript ± refers to the helicities ±1/2. TheDirac equation then reduces to an equation with no derivatives:

Eψ = (α · ~p+mβ)ψ, (8)

where ~p is the momentum vector (not an operator). Explicit solutions can beobtained easily as

u+(p) =1√

E +m

((E +m)χ+(~p)

pχ+(~p)

), (9)

u−(p) =1√

E +m

((E +m)χ−(~p)−pχ−(~p)

). (10)

Here and below, we adopt normalization u†±(p)u±(p) = 2E and E =√~p2 +m2.

Negative energy solutions must be filled in the vacuum and their “holes” areregarded as anti-particle states. Therefore, it is convenient to assign momentum−~p and energy −E = −

√~p2 +m2. The solutions have space and time dependence

ψ±(x, t) = v±(p)e+iEt−i~p·~x. The Dirac equation again reduces to an equation withno derivatives:

−Eψ = (−α · ~p+mβ)ψ. (11)

Explicit solutions are given by

v+(p) =1√

E +m

(−pχ−(~p)

(E +m)χ−(~p)

), (12)

v−(p) =1√

E +m

(pχ+(~p)

(E +m)χ+(~p)

). (13)

It is convenient to define “barred” spinors u = u†γ0 = u† and v = v†γ0. Thecombination uu is a Lorentz-invariant, uu = 2m, and similarly, vv = −2m. Thecombination uγµu transforms as a Lorentz vector:

uκ(p)γµuλ(p) = 2pµδκ,λ, (14)

where κ, λ = ±, and similarly,

vκ(p)γµvλ(p) = 2pµδκ,λ. (15)

They can be interpreted as the “four-current density” which generates electromag-netic field: uγ0u = u†u is the “charge density,” and uγiu = u†αiu is the “currentdensity.”

Note that the matrix

γ5 = iγ0γ1γ2γ3 =

(0 11 0

)(16)

commutes with the Hamiltonian in the massless limit m → 0. In fact, at highenergies E � m, the solutions are almost eigenstates of γ5, with eigenvalues +1for u+ and v−, and −1 for u− and v+. The eigenvalue of γ5 is called “chirality.”Therefore chirality is a good quantum number in the high energy limit. Neutrinoshave chirality minus, and they do not have states with positive chirality.